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3 years ago

7. A box is pulled straight across the floor at a constant speed. It is pulled with a horizontal force of 48 N.

Physics

1 answer:

Leya [2.2K]3 years ago

4 0

Answer:

a)Taking into consideration Newton’s second law, we know that

Net_Force = mass * acceleration

Since the box is pulled at constant speed, the acceleration is equal to zero.

This means that

Net_Force = 0 N

b) Force of friction

The net force is equal to the sum of all forces,

Net_Force = Force_applied - Friction

We found that Net_Force = 0, which means

Friction = Force_applied = 48 N

c) If the box comes to a stop. And the applied force becomes zero, the friction force becomes also zero.

Friction = 0 N

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What is newton's second law of motion?

lord [1]

Answer:

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

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3 years ago

Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge

Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃= 2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

$\frac{k*q_{1}*q_3 }{(d_{13})^{2} } = \frac{k*q_{2}*q_3 }{(d_{23})^{2} }$

We divide by (k * q3) on both sides of the equation

$\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }$

$q_{1} = \frac{q_{2}*(d_{13})^{2} }{(d_{23} )^{2} }$

$q_{1} = \frac{5*(2)^{2} }{(4 )^{2} }$

q₁ = + 1.25 nC

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3 years ago

Plan an experiment to measure the ideal mechanical advantage of a three-hole punch. (a) What materials would you need? (b) What

Vaselesa [24]

Answer:

A) Three hole punch and either a layered plastic or paper

B) Identify the lengths involved ,

Length of input arm / length of output arm = L1/ L2

Explanation:

<u>a) Materials involved includes :</u>

Three hole punch and either a layered plastic or paper

Identify the forces acting on the three-hole punch which are Input and output forces

Identify the points where they act

<u>B) procedures involved </u>

The mechanical advantage = output force / input force

step one: Identify the lengths involved

assuming no friction or relatively small friction \

mechanical advantage can be calculated as : Length of input arm / length of output arm = L1/ L2

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3 years ago

A blank is any disturbance that carries energy from one place to another through matter and space.

Eduardwww [97]

<h3>Answer;</h3>

<em><u>A wave</u></em> is any form of a disturbance that carries energy from one place to another through a matter and space

<h3>Explanation;</h3>

Waves carry energy from one point, the source to another point or place. The transmission of a wave may occur through the space or through a material medium.
Electromagnetic waves are those waves whose transmissions occurs through the space, they do not require material medium for transmission,for example, radio waves, while mechanical waves are those that require material medium for transmission, for example sound waves.
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3 years ago

Read 2 more answers

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with

Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = $\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2$

= $\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2$

= 31 J

Final KE = $\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J$

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0

3 years ago